c
When the given metallic spheres are connected by a conducting wire, charge will flow till both the spheres acquire a common potential which is given by Common potential,

\(V =\frac{q_{1}+q_{2}}{C_{1}+C_{2}}=\frac{-1 \times 10^{-2}+5 \times 10^{-2}}{4 \pi \varepsilon_{0} R_{1}+4 \pi \varepsilon_{0} R_{2}}\)

\(=\frac{4 \times 10^{-2}}{4 \pi \varepsilon_{0}\left(1 \times 10^{-2}+3 \times 10^{-2}\right)}\)

\(=\frac{4 \times 10^{-2}}{4 \pi \varepsilon_{0} \times 4 \times 10^{-2}}.........(i)\)

\(\therefore \) Final charge on the bigger sphere 

\(=4 \pi \varepsilon_{0} \times 3 \times 10^{-2} \times \frac{4 \times 10^{-2}}{4 \pi \varepsilon_{0} \times 4 \times 10^{-2}}(\mathrm{Using}(\mathrm{i}))\)

\(=3 \times 10^{-2} \mathrm\,{C}\)