Answer
a) ${\bf 12.81}\;\rm \mu s$
b) ${\bf 0.625} \; c$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the spacetime interval between the two events is invariant in all frames.
So
$$s^2=(c\Delta t)^2-( \Delta x)^2=(c\Delta t')^2-( \Delta x')^2$$
$$ c^2(\Delta t)^2-( \Delta x)^2=c^2(\Delta t')^2-( \Delta x')^2$$
Solving for $( \Delta t')$
$$ (\Delta t')^2= \dfrac{c^2(\Delta t)^2-( \Delta x)^2+( \Delta x')^2}{c^2} $$
$$ (\Delta t')=\sqrt{ \dfrac{c^2(\Delta t)^2-( \Delta x)^2+( \Delta x')^2}{c^2} }$$
Plug the known,
$$ (\Delta t')=\sqrt{ \dfrac{(3\times 10^8)^2(10\times 10^{-6})^2-( 0)^2+( 2400)^2}{(3\times 10^8)^2} }$$
$$ (\Delta t')=\color{red}{\bf 12.81}\;\rm \mu s$$
$$\color{blue}{\bf [b]}$$
Using Lorentz transformation for time.
$$t'=\gamma\left[t-\dfrac{vx}{c^2}\right] =\gamma\left[t-\dfrac{v(0)}{c^2}\right] =\gamma \;t$$
where $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2} $
Hence,
$$t'=\left[1-\frac{v^2}{c^2}\right]^{-1/2} \;t$$
$$\dfrac{t}{t'}= \sqrt{1-\frac{v^2}{c^2}} $$
Squaring both sides;
$$\dfrac{t^2}{(t')^2}= 1-\frac{v^2}{c^2} $$
$$ v^2 = c^2\left[1- \dfrac{t^2}{(t')^2} \right]$$
$$ v = c\;\sqrt{\left[1- \dfrac{t^2}{(t')^2} \right]}$$
Plug the known;
$$ v = c\;\sqrt{\left[1- \dfrac{(10\mu)^2}{(12.81)^2} \right]}$$
$$ v =\color{red}{\bf 0.625} \; c$$
