c
(c) \(Q\) = \(\sigma\) \(A t\) (\(T_4\) -\(T_0^4\)) 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

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