Answer
See the detailed answer below.
Work Step by Step
$$\bf a)$$ We know that the speed of a wave is given by
$$v=\lambda f=\dfrac{\lambda}{T}$$
where $f$ is the frequency and $T$ is the periodic time.
Hence,
$$\lambda=vT=(4)(0.2)=\color{red}{\bf0.8 }\;\rm m$$
$$\bf b)$$
We know that the function of this wave is given by
$$D_{(x,t)}=A\sin\left[ \dfrac{2\pi x}{\lambda}-\dfrac{2\pi t}{T}+\phi_0 \right]$$
And at $t=0$, $x=0$,
$$D_{(0,0)}=A\sin\left[ \dfrac{2\pi (0)}{\lambda}-\dfrac{2\pi (0)}{T}+\phi_0 \right]=A\sin\phi_0$$
where at $t=0$, $D=A$, so
$$\phi_0=\sin^{-1}(1)=\color{red}{\bf \dfrac{\pi}{2}}\;\rm rad$$
$$\bf c)$$
The displacement equation for the wave is given by
$$D_{(x,t)}=A\sin\left[ \dfrac{2\pi x}{\lambda}-\dfrac{2\pi t}{T}+\phi_0 \right]$$
Plugging the known;
$$D_{(x,t)}=(2\times 10^{-3})\sin\left[ \dfrac{2\pi }{0.8}(x)-\dfrac{2\pi }{0.2}(t)+\dfrac{\pi}{2}\right]$$
$$\boxed{D_{(x,t)}=(0.002)\sin\left[ 2.4\pi x -10\pi t+\dfrac{\pi}{2}\right]}$$
