Answer
(a) increased by $\sqrt 2$
(b) $15813m/s$
(c) same
Work Step by Step
(a) We know that
$v_e=\sqrt{\frac{2GM_E}{R_E}}$
Given that the mass is doubled then
$v_e=\sqrt{\frac{2G(2M_E)}{R_E}}$
$\implies v_e=\sqrt{\frac{4GM_E}{R_E}}$
Thus, the escape speed is increased by a factor of $\sqrt 2$.
(b) We can calculate the required escape speed as
$v_e=\sqrt{\frac{4GM_E}{R_E}}$
We plug in the known values to obtain:
$v_e=\sqrt{\frac{4(6.67\times 10^{-11}N.m^2/Kg^2)(5.97\times 10^{24}Kg)}{6.37\times 10^6m}}$
$v_e=15813m/s$
(c) We know that $v_e=\sqrt{\frac{2GM_E}{R_E}}$. This equation shows that the escape speed depends on mass and not on the mass of the rocket. Thus, the escape speed remains the same.