Answer
$L = 0.357~m$
Work Step by Step
We can find the period as:
$T = \frac{time}{oscillations}$
$T = \frac{12.0~s}{10~oscillations}$
$T = 1.20~s$
We then find the length of the string as:
$T = 2\pi~\sqrt{\frac{L}{g}}$
$L = \frac{T^2~g}{(2\pi)^2}$
$L = \frac{(1.20~s)^2~(9.80~m/s^2)}{(2\pi)^2}$
$L = 0.357~m$
