Answer
$$x_{g}=2.6 \mathrm{\ m}$$
Work Step by Step
it's obvious that accelerations of the girl and sled are in opposite directions.
Assuming the girl starts at the origin and moves in the $+x$ direction, her coordinate is given by $x_{g}=\frac{1}{2} a_{g} t^{2} .$
The sled starts at $x_{0}=15 \mathrm{m}$ and moves in the $-x$ direction. Its coordinate is given by $x_{s}=x_{0}-\frac{1}{2} a_{s} t^{2} .$
They meet when $x_{g}=x_{s},$ or
$$\frac{1}{2} a_{g} t^{2}=x_{0}-\frac{1}{2} a_{s} t^{2}$$
This occurs at time
$$t=\sqrt{\frac{2 x_{0}}{a_{g}+a_{s}}}$$
By then, the girl has gone the distance
$$x_{g}=\frac{1}{2} a_{g} t^{2}=\frac{x_{0} a_{g}}{a_{g}+a_{s}}=\frac{(15 \mathrm{m})\left(0.13 \mathrm{m} / \mathrm{s}^{2}\right)}{0.13 \mathrm{m} / \mathrm{s}^{2}+0.62 \mathrm{m} / \mathrm{s}^{2}}=2.6 \mathrm{m}$$
