Answer
$2\;\rm kg$
Work Step by Step
In the first case, we know that the weight of the hanging mass is equal to the tension in the wire.
So,
$$T_s=Mg\tag 1$$
And in the second case,
$$T_s'=(M+m)g\tag 2$$
Recall that the second-harmonic frequency, in the closed-closed tube which is similar to the two-fixed-ends wire here, is given by
$$f_m=\dfrac{mv}{2L}$$
where $v=\sqrt{T_s/\mu}$
In the first case,
$$f_2=\dfrac{2}{2L}\sqrt{\dfrac{T_s}{\mu}}$$
Plugging from (1);
$$f_2=\dfrac{1}{L}\sqrt{\dfrac{Mg}{\mu}}\tag 3$$
And in the second case, using (2)
$$f_2'=\dfrac{1}{L}\sqrt{\dfrac{(M+m)g}{\mu}}\tag 4$$
Divide (4) by (3);
$$\dfrac{f_2'}{f_2}=\dfrac{\dfrac{1}{L}\sqrt{\dfrac{(M+m)g}{\mu}}}{\dfrac{1}{L}\sqrt{\dfrac{Mg}{\mu}}}= \sqrt{\dfrac{M+m}{M}}$$
Squaring both sides;
$$\left[\dfrac{ f_2' }{f_2 } \right]^2 =\dfrac{M+m}{M} =1+\dfrac{m}{M}$$
Hence,
$$M=\dfrac{m}{\left[\dfrac{ f_2' }{f_2 } \right]^2-1}$$
Plugging the known;
$$M=\dfrac{1.0}{\left[\dfrac{ 245 }{200} \right]^2-1}$$
$$M=\color{red}{\bf 2.0}\;\rm kg$$
