If $T, T_0, \sigma $ and t are same for both bodies then $\frac{{{Q_{sphere}}}}{{{Q_{cube}}}} = \frac{{{A_{sphere}}}}{{{A_{cube}}}} = \frac{{4\pi {r^2}}}{{6{a^2}}}$ …..$(i)$

But according to problem, volume of sphere = Volume of cube

==> $\frac{4}{3}\pi {r^3} = {a^3}$

==> $a = {\left( {\frac{4}{3}\pi } \right)^{1/3}}r$

Substituting the value of a in equation $(i)$ we get

$\frac{{{Q_{sphere}}}}{{{Q_{cube}}}} = \frac{{4\pi {r^2}}}{{6{a^2}}} = \frac{{4\pi {r^2}}}{{6{{\left\{ {{{\left( {\frac{4}{3}\pi } \right)}^{1/3}}r} \right\}}^2}}}$

$ = \frac{{4\pi {r^2}}}{{6\,{{\left( {\frac{4}{3}\pi } \right)}^{2/3}}{r^2}}} = {\left( {\frac{\pi }{6}} \right)^{1/3}}:1$