Recall that a perfect square is an integer whose square root is also an integer. For example, the square root of the integer 36 is the integer 6, so 36 is a perfect square.
The important thing to remember about questions like these is that the way the question is worded implies that the parity of the tens digit is the same for all x that meet the criteria. If this were not the case, then the question would have more than one correct answer! (It would be a trick question.) Since the question does not demand a proof, then we only have to find one example of a perfect square x where the units (ones) digit is 4. Then, the parity of the tens digit of x gives us our answer.
Recall that the parity of an integer is the evenness or oddness of the number. For example, the parity of 2 is even. We can simply choose x = 4, because the square root of 4 is the integer 2. In this case, the invisible tens digit of x is 0. Since 0 is an even number, we have our answer: The parity of the tens digit of x is even.
Here is the proof for all perfect squares with units digit 4, if you're interested or if it's required.
Let's approach this by obtaining x by squaring its square roots. In fact you can generate all the perfect squares by squaring every whole number in turn: 0 → 0, 1 → 1, 2 → 4, 3 → 9, 4 → 16, etc. We need to narrow down the possibilities for these square roots somehow.
Let's consider which square roots produce perfect squares that have 4 as the units digit. When you multiply any two integers, if you know the units digits of the two factors, then you know the units digit of the product: It's simply the product of the units digits of the factors. For instance, the product of 123 and 456 has units digit 8, because 3 times 6 is 18, which has units digit 8. When you list all of the perfect squares corresponding to the square roots 0 through 9, you are in fact listing all of the possibilities for the units digit of a perfect square. You should be able to determine that the only square roots that yield perfect squares with units digit 4 are square roots with units digit 2 or 8.
When we square a generic three-digit number (d2•102 + d1•10 + d0) with digits (d2, d1, d0), watch what happens:
(d2•100 + d1•10 + d0)2 = (d2•100)(d2•100) + (d2•100)(d1•10) + (d2•100)(d0) + (d1•10)(d2•100) + (d1•10)(d1•10) + (d1•10)(d0) + (d0)(d2•100) + (d0)(d1•10) + (d0)(d0)
= d22•10000 + d2d1•1000 + d2d0•100 + d1d2•1000 + d12•100 + d1d0•10 + d0d2•100 + d0d1•10 + d02
I bolded the only terms above that can possibly determine the tens digit of the perfect square. When we combine these terms, we get 20d0d1 + d02. Thus, the only digits of the square root that are involved in the calculation of the tens place are d0 and d1, which are the units and tens place, respectively, of the square root. Even though we chose a three digit square root as an example, this fact is true for integers having more digits, as well.
Now remember we said that the units digit of the square root can only be 2 or 8. When we plug in d0 = 2, we get 20d0d1 + d02 = 20(2)d1 + 22 = 40d1 + 4. Since the 4 term doesn't affect the resulting tens digit, and no matter the value of d1 the tens digit is even, we have that the tens digit of the perfect square is even when the units digit of the square root is 2. When we plug in d0 = 8, we get 20d0d1 + d02 = 20(8)d1 + 82 = 160d1 + 60 + 4. Again, no matter the value of d1, this perfect square's tens digit is even, too.
Thus, we have shown that, when the units digit of a perfect square is 4, its tens digit is even.