Решение.\[ \begin{align}
& I={{I}_{A}}+{{I}_{R}}\ \ \ (1). \\
& {{U}_{A}}={{I}_{A}}\cdot {{R}_{A}}\ \ \ (2).\ {{U}_{R}}={{U}_{A}}\ \ \ (3).\ R=\rho \cdot \frac{l}{S}\ \ \ (4).\ S=\frac{\pi \cdot {{d}^{2}}}{4}\ \ \ (5). \\
& {{I}_{R}}=\frac{{{U}_{R}}}{R}\ \ \ (6).\ {{I}_{R}}=\frac{{{I}_{A}}\cdot {{R}_{A}}\cdot \pi \cdot {{d}^{2}}}{\rho \cdot l\cdot 4}\ \ \ (7). \\
& I={{I}_{A}}+\ \frac{{{I}_{A}}\cdot {{R}_{A}}\cdot \pi \cdot {{d}^{2}}}{\rho \cdot l\cdot 4}\ . \\
\end{align} \]
I = 13,3 А.