a) 

Utilizando a equação 2.43 temos:

\(\begin{align} & V=\frac{q}{8\pi \varepsilon R}\left( 3-\frac{{{r}^{2}}}{{{R}^{2}}} \right) \\ & W=\frac{1}{2}\int_{0}^{R}{\rho V}d\tau \\ & W=\frac{q\rho }{4\varepsilon R}\left[ \frac{3{{r}^{3}}}{3}-\frac{{{r}^{5}}}{5{{R}^{2}}} \right]_{0}^{R} \\ & W=\frac{1}{4\pi \varepsilon }\left( \frac{3{{q}^{2}}}{5R} \right) \\ \end{align}\ \)

\(\boxed{W = \frac{1}{{4\pi \varepsilon }}\left( {\frac{{3{q^2}}}{{5R}}} \right)}\)

b) Usando a equação 2.45 temos:

\(\begin{align} & W=\frac{\varepsilon }{2}\int_{{}}^{{}}{{{E}^{2}}d\tau } \\ & W=\frac{\varepsilon }{2}\left( \frac{{{q}^{2}}}{{{\left( 4\pi \varepsilon \right)}^{2}}} \right)\left[ \int_{0}^{R}{{{\left( \frac{r}{{{R}^{3}}} \right)}^{2}}\left( {{r}^{2}}4\pi \right)dr+\int_{R}^{\propto }{\frac{1}{{{r}^{4}}}\left( {{r}^{2}}4\pi \right)dr}} \right] \\ & W=\frac{{{q}^{2}}}{\left( 8\pi \varepsilon \right)}\left( \frac{1}{R}+\frac{1}{5R} \right) \\ & W=\frac{1}{\left( 4\pi \varepsilon \right)}\left( \frac{3{{q}^{2}}}{5R} \right) \\ \end{align}\ \)

\(\boxed{W = \frac{1}{{4\pi \varepsilon }}\left( {\frac{{3{q^2}}}{{5R}}} \right)}\)

c) Usando a equação 2.44 temos:

\(\begin{align} & W=\frac{\varepsilon }{2}\left( \int_{{}}^{{}}{{{E}^{2}}d\tau }+\int_{{}}^{{}}{VEda} \right) \\ & W=\frac{\varepsilon }{2}\left( \frac{{{q}^{2}}4\pi }{a{{(4\pi \varepsilon )}^{2}}}+\frac{{{q}^{2}}4\pi }{5R{{(4\pi \varepsilon )}^{2}}}+\frac{{{q}^{2}}4\pi }{{{(4\pi \varepsilon )}^{2}}}\left( \frac{-1}{r} \right) \right) \\ & W=\frac{{{q}^{2}}4\pi }{8(4\pi \varepsilon )}\left( \frac{1}{a}+\frac{1}{5R}-\frac{1}{a}+\frac{1}{R} \right) \\ & W=\frac{1}{(4\pi \varepsilon )}\left( \frac{3{{q}^{2}}}{5R} \right) \\ \end{align}\ \)

\(\boxed{W = \frac{1}{{4\pi \varepsilon }}\left( {\frac{{3{q^2}}}{{5R}}} \right)}\)