If x=years since 1996 and y=# of childbirths, then our data points are (3,13), (5,25), (9,47), (10,51), and (13,72), and we wish to fit the power function y=axb to this data.
This question is easier to solve by transforming it using logarithms, and first solving the problem of fitting the function ln(y) = ln(axb) to the data (ln(3),ln(13)), (ln(5),ln(25)), (ln(9),ln(47)), (ln(10),ln(51)), and (ln(13),ln(72)).
Using properties of logarithms, we rewrite the function:
ln(y) = ln(axb)
ln(y) = ln(a) + ln(xb)
ln(y) = ln(a) + b·ln(x)
If we let Y=ln(y), A=ln(a), and X=ln(x), then we have the linear function Y=A+bX.
Now we can do linear regression and fit the line of least squares to the data. You can use a calculator (using STAT>CALC>LinReg(a+bx) on a TI-84), Excel (using =LINEST()), or similar software, or do the calculations manually by using the formulas
slope: b=(n∑xiyi – ∑xi∑yi)/(n∑xi2 – (∑xi)2)
Y-intercept: A = (∑yi – b∑xi)/n
where n is the number of data points (so n=5 in this question).
We find b=1.1441 and A=1.3323. Now we have the line Y=1.3323+1.1441X fit to the data
(ln(3),ln(13)), (ln(5),ln(25)), (ln(9),ln(47)), (ln(10),ln(51)), and (ln(13),ln(72)).
To return to the original question, we undo the logarithm by applying its inverse:
eY = e1.3323+1.1441X
eln y = e1.3323e1.1441ln x
y = e1.3323eln x^1.1441
y = 3.7897x1.1441
Now this power function fits the original data (3,13), (5,25), (9,47), (10,51), and (13,72).
2013 is 17 years after 1996. With x=17, we get a prediction of y = 3.7897·171.1441 ≈ 97 childbirths.