Answer
The speed will be $\frac{v_0}{2}$
Work Step by Step
Let $m_1$ be the mass of the first string;
$m_1 = \rho~V_1$
$m_1 = \rho~(L~\pi~R^2)$
We can find an expression for the mass of the new string as:
$m_2 = \rho~V_1$
$m_2 = \rho~L~\pi~(2R)^2$
$m_2 = 4~\rho~(L~\pi~R^2)$
$m_2 = 4~m_1$
We can find an expression for the original speed as:
$v_0 = \sqrt{\frac{F_T}{\mu}}$
$v_0 = \sqrt{\frac{F_T}{m_1/L}}$
$v_0 = \sqrt{\frac{F_T~L}{m_1}}$
We can find an expression for the new speed $v_2$ as:
$v_2 = \sqrt{\frac{F_T}{\mu}}$
$v_2 = \sqrt{\frac{F_T}{m_2/L}}$
$v_2 = \sqrt{\frac{F_T~L}{m_2}}$
$v_2 = \sqrt{\frac{F_T~L}{4~m_1}}$
$v_2 = \frac{1}{2}\sqrt{\frac{F_T~L}{m_1}}$
$v_2 = \frac{1}{2}~v_0$
The speed will be $\frac{v_0}{2}$.
