Answer
$2.40\times 10^4Km$
Work Step by Step
We can find the required distance at apogee as follows:
$\frac{1}{r_a}=\frac{1}{r_p}+\frac{v_a^2-v_p^2}{2GM}$
We plug in the known values to obtain:
$\frac{1}{r_a}=\frac{1}{2.00\times 10^7}+\frac{(3640)^2-(4460)^2}{2(6.67\times 10^{-11})(5.97\times 10^{24})}$
$\frac{1}{r_a}=4.17\times 10^{-8}m^{-1}$
$\implies r_{a}=\frac{1}{4.17\times 10^{-8}}$
$r_{a}=2.40\times 10^7m=2.40\times 10^4Km$
