Answer
$8.02\;\rm m/s^2$
Work Step by Step
When the wire is horizontal and attached to the pulley, there will be a standing wave on the horizontal part of the wire.
The frequency is given by
$$f_m=\dfrac{mv}{2L}=\dfrac{m}{2L}\sqrt{\dfrac{T_s}{\mu}}$$
where $L$ here is $\frac{1}{2}L_{wire}$
And for the second-harmonic frequency
$$f_2=\dfrac{2}{2L}\sqrt{\dfrac{T_s}{\mu}}$$
Plugging the known and solving for $T_s$
$$100=\dfrac{1}{\frac{1}{2}L_{wire}}\sqrt{\dfrac{T_s}{\mu}}$$
$$100=\dfrac{2}{L_{wire}}\sqrt{\dfrac{T_s}{\mu}}\tag 1$$
We know that the tension in the wire is equal to the weight of the hanging mass since we assume that the pulley is frictionless.
Thus,
$$\sum F_y=T_s-Mg'=0$$
where $g'$ is the free-fall acceleration in the Physics planet.
$$T_s=Mg'$$
Plugging into (1);
$$100=\dfrac{2}{L_{wire}}\sqrt{\dfrac{Mg'}{\mu}} $$
$$50L_{wire}=\sqrt{\dfrac{Mg'}{\mu}} $$
Squaring both sides:
$$50^2L_{wire}^2=\dfrac{Mg'}{\mu} $$
Thus,
$$g'M= 2500\mu L_{wire}^2 \tag 2$$
Now we need to find the length of the wire.
We know for a simple pendulum that the period is given by
$$T=2\pi\sqrt{\dfrac{L_{wire}}{g'}}$$
where $T$ is the period.
Hence,
$$L_{wire}=\dfrac{T^2g'}{4\pi^2}$$
Plugging the known;
$$L_{wire}=\dfrac{\left(\dfrac{314}{100}\right)^2g'}{4\pi^2}$$
Plugging into (2);
$$g'M= 2500\mu\left[\dfrac{\left(\dfrac{314}{100}\right)^2g'}{4\pi^2}\right]^2 $$
$$ \color{red}{\bf\not} g'M= 2500\mu \dfrac{\left(\dfrac{314}{100}\right)^4(g')^{ \color{red}{\bf\not} 2}}{16\pi^4} $$
Thus,
$$g'=\dfrac{16\pi^4M}{2500\mu\left(\dfrac{314}{100}\right)^4}$$
Plugging the known;
$$g'=\dfrac{16\pi^4(1.25)}{2500(1\times 10^{-3})\left(\dfrac{314}{100}\right)^4}$$
$$g'=\color{red}{\bf 8.02}\;\rm m/s^2$$
