Answer
$a= 0.07292 \space m^2 = 7.3 \times 10^{-2} \space m^2$
Work Step by Step
$R= \frac{ln 2}{T_{1/2}} N $
$N = (\frac{M}{m}) (\frac{a}{A})$
Substitute N into the equation
$R= \frac{ln 2}{T_{1/2}} (\frac{M}{m}) (\frac{a}{A})$
To find the ground area, solve for a. Rearrange the equation,
$a= \frac{AmRT_{1/2} }{M ln 2}$
Where
M is the mass of $^{90}Sr$ produced = $400g$
m is the mass of a $^{90}Sr$ nucleus = $\frac{90 g/mol}{6.02 \times 10^{23}/mol} = 1.495 \times 10^{-22} g$
A is the area of radioactive fallout $ = 2000 km^2 = 2.0 \times 10^9 m^2$
Half life of the nucleus $T_{1/2} = 29 y \times 3.15 \times 10^7 s/y = 9.135 \times 10^8 s $
So,
$a= \frac{(2.0 \times 10^9 m^2)(1.495 \times 10^{-22} g)(74000/s)(9.135 \times 10^8 s ) }{(400g) (0.693)}$
$a= 0.07292 \space m^2 = 7.3 \times 10^{-2} \space m^2$
