Answer
$\approx {\bf 46}\;\rm m/s$
Work Step by Step
Let's assume that the frame of the stationary clock is $S$, and the frame of the moving one is $S'$.
We are given that the time difference for the moving clock is
$$dt' =1\;\rm day=\bf (86400)\;\rm s\tag 1$$
while then the time difference for the stationary clock
$$ dt = \rm1.0 \;day+1\;ns =\bf ( 86400+10^{-9}) \rm\;s\tag 2$$
According to time dilation, a moving clock runs slower than a stationary clock. So we can use the following equation for time relation.
$$dt=\gamma \;dt' \tag 3$$
where $\gamma=1/\sqrt{1-\frac{v^2}{c^2}}$
As the author told us, we can use the binomial approximation for $\gamma$,
$$\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}\approx \left[1-\frac{-1}{2}\dfrac{v^2}{c^2}\right]$$
$$\gamma\approx 1+\dfrac{v^2}{2c^2}$$
Plug into (3);
$$dt= \left[ 1+\dfrac{v^2}{2c^2}\right]dt'$$
Solving for $v$;
$$\dfrac{dt}{dt'}= 1+\dfrac{v^2}{2c^2} $$
$$\dfrac{dt}{dt'}-1= \dfrac{v^2}{2c^2} $$
$$v=c\;\sqrt{2\left[\dfrac{dt}{dt'}-1\right]} $$
Plug the known and from (1), and (2);
$$v=(3\times 10^8) \sqrt{2\left[\dfrac{( 86400+10^{-9})}{(86400 )}-1\right]} $$
$$v=45.6\;\rm m/s\approx \color{red}{\bf 46}\;\rm m/s$$
