Answer
a) $1.12\times 10^{-20}\;\rm J$
b) $1 .8\times 10^{21}\;\rm ion$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The work needed to move one sodium ion from the inside of the cell to the outside is given by
$$W=q \Delta V=e\Delta V$$
where the net charge of one sodium ion is $q=+e$
Plug the known,
$$W=(1.6\times 10^{-19})(70\times 10^{-3})$$
$$W=\color{red}{\bf 1.12\times 10^{-20}}\;\rm J$$
$$\color{blue}{\bf [b]}$$
We need energy of $ 1.12\times 10^{-20}\;\rm J$ to move one ion, and the body uses 20% of 100 J to move sodium ions, which is about 20 J.
Hence, the number of ions is given by
$$N=\dfrac{W_{tot}}{W_{ion}}=\dfrac{20}{1.12\times 10^{-20}}$$
$$N=\color{red}{\bf1 .8\times 10^{21}}\;\rm ion$$
