Answer
$\frac{mv}{\sqrt{{{m}^{2}}-{{m}_{0}}^{2}}}$.
Work Step by Step
$m=\frac{{{m}_{0}}}{\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}$
Multiply by$\left( \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \right)$ on both sides of the equation,
$\begin{align}
& m=\frac{{{m}_{0}}}{\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}} \\
& \left( m \right)\times \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}=\frac{{{m}_{0}}}{\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}\times \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \\
& m\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}={{m}_{0}}
\end{align}$
Divide by $\left( m \right)$ on both sides,
$\begin{align}
& \frac{m\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}{m}=\frac{{{m}_{0}}}{m} \\
& \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}=\frac{{{m}_{0}}}{m}
\end{align}$
Squaring both sides,
$\begin{align}
& \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}=\frac{{{m}_{0}}}{m} \\
& {{\left( \sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}} \right)}^{2}}={{\left( \frac{{{m}_{0}}}{m} \right)}^{2}} \\
& 1-\frac{{{v}^{2}}}{{{c}^{2}}}=\frac{{{m}_{0}}^{2}}{{{m}^{2}}}
\end{align}$
Multiply by ${{c}^{2}}{{m}^{2}}$ on both sides of the equation.
$\begin{align}
& \left( 1-\frac{{{v}^{2}}}{{{c}^{2}}} \right)\times {{c}^{2}}{{m}^{2}}=\frac{{{m}_{0}}^{2}}{{{m}^{2}}}\times {{c}^{2}}{{m}^{2}} \\
& {{c}^{2}}{{m}^{2}}-{{m}^{2}}{{v}^{2}}={{c}^{2}}{{m}_{0}}^{2} \\
& {{c}^{2}}\left( {{m}^{2}}-{{m}_{0}}^{2} \right)={{m}^{2}}{{v}^{2}}
\end{align}$
Divide by ${{m}^{2}}-{{m}_{0}}^{2}$ on both sides of the equation,
$\begin{align}
& {{c}^{2}}\left( {{m}^{2}}-{{m}_{0}}^{2} \right)={{m}^{2}}{{v}^{2}} \\
& \frac{{{c}^{2}}\left( {{m}^{2}}-{{m}_{0}}^{2} \right)}{{{m}^{2}}-{{m}_{0}}^{2}}=\frac{{{m}^{2}}{{v}^{2}}}{{{m}^{2}}-{{m}_{0}}^{2}} \\
& {{c}^{2}}=\frac{{{m}^{2}}{{v}^{2}}}{{{m}^{2}}-{{m}_{0}}^{2}}
\end{align}$
Considering only the positive square root:
$\begin{align}
& {{c}^{2}}=\frac{{{m}^{2}}{{v}^{2}}}{{{m}^{2}}-{{m}_{0}}^{2}} \\
& c=\frac{mv}{\sqrt{{{m}^{2}}-{{m}_{0}}^{2}}}
\end{align}$
Thus, the required value is $c=\frac{mv}{\sqrt{{{m}^{2}}-{{m}_{0}}^{2}}}$.
