Answer
a) $1.375\times 10^5\;\rm m$
b) $8.64\times 10^6\;\rm m/s$
Work Step by Step
a) We can say that the angular momentum of the star is conserved.
$$L_i=L_f$$
where $L_i$ is the star's angular momentum before it collapses and $L_f$ is its angular momentum after it collapses to a neutron star.
$$I_i\omega_i=I_f\omega_f$$
We know that $\omega=2\pi/T$ where $T$ is the periodic time it takes to complete one full cycle.
Thus,
$$I_i\dfrac{ \color{red}{\bf\not}2 \color{red}{\bf\not}\pi}{T_i}=I_f\dfrac{ \color{red}{\bf\not}2 \color{red}{\bf\not}\pi}{T_f}$$
Hence,
$$\dfrac{I_iT_f}{T_i}=I_f $$
we know that the star can be represented as a solid sphere, so its moment of inertia around its center of mass is given by $I=\frac{2}{5}MR^2$.
So that;
$$\dfrac{ \color{red}{\bf\not}\frac{2}{5} \color{red}{\bf\not}MR_i^2T_f}{T_i}= \color{red}{\bf\not}\frac{2}{5} \color{red}{\bf\not}MR_f^2$$
Noting that the mass of the star is constant but its radius decreases.
$$\dfrac{ R_i^2T_f}{T_i}= R_f^2$$
Therefore,
$$R_f=\sqrt{\dfrac{ T_f}{T_i}} R_i$$
We know that initially, the star was completing one full cycle every 30 days (which is $T_i$) and after the collapse, it is completing one full cycle every 0.1 seconds (which is $T_f$).
$$R_f=\sqrt{\dfrac{ 0.1}{30\times 24\times 60\times 60}} \times (7\times 10^8)$$
$$R_f=\color{red}{\bf 1.375\times 10^5}\;\rm m$$
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b) We can find the speed of a point on the equator of our neutron star by dividing the circumference at the equator by the periodic time.
$$v=\dfrac{2\pi R_f}{T_f}=\dfrac{2\pi \times1.375\times 10^5 }{0.1}$$
$$v=\color{red}{\bf 8.64\times 10^6}\;\rm m/s$$
