The escape velocity on earth’s surface is, \[v_e=\sqrt{2gR}\]
The escape velocity on the surface of another planet is, \[v_e’=\sqrt{2g’R’}\]
According to the question, $M’=\frac{M}{4}$ and $R’=\frac{R}{3}$.
\[\therefore g’=\frac{GM’}{R’^2}=\frac{G(M/4)}{(R/3)^2}\] \[=\frac{9}{4}\left(\frac{GM}{R^2}\right)=\frac{9}{4}g\]
Now, \[v_e’=\sqrt{2\frac{9}{4}g\frac{R}{3}}=\sqrt{\frac{3}{4}}\sqrt{2gR}=\sqrt{\frac{3}{4}}v_e\]
Putting $v_e=11.2$ km/sec, we get, \[v_e’=\sqrt{\frac{3}{4}}×11.2=9.7\;\text{km/sec}\]