Solution:

Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$. Since mass is constant and radius becomes one-fourth, $v_e$ increases by a factor of $\sqrt{4} = 2$. Thus, $v_p = 2 \times 11.2\text{ km/s} = 22.4\text{ km/s}$. Option B is correct.