Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need to find the phase difference in both cases, right and left.
$$D_{1,\rm Right}=a\sin\left[ 2\pi\left(\dfrac{x}{\lambda}-\dfrac{t}{T}\right)\right]$$
$$D_{2,\rm Right}=a\sin\left[ 2\pi\left(\dfrac{x-L}{\lambda}-\dfrac{t}{T}\right)+\phi_{20}\right]$$
Hence, the phase difference between them on the right side is given by
$$\Delta \phi_{\rm Right}=\phi_2-\phi_1$$
$$\Delta \phi_{\rm Right}= 2\pi\left(\dfrac{x-L}{\lambda}-\dfrac{t}{T}\right)+\phi_{20}-\left[ 2\pi\left(\dfrac{x}{\lambda}-\dfrac{t}{T}\right)\right]$$
$$\Delta \phi_{\rm Right}= 2\pi\left(\dfrac{x-L}{\lambda}-\dfrac{t}{T}\right)+\phi_{20}- 2\pi\left(\dfrac{x}{\lambda}-\dfrac{t}{T}\right) $$
$$\Delta \phi_{\rm Right}= 2\pi\left[ \dfrac{x-L}{\lambda}-\dfrac{t}{T} - \dfrac{x}{\lambda}+\dfrac{t}{T} \right]+\phi_{20}$$
$$\Delta \phi_{\rm Right}= 2\pi\left[ \dfrac{x-L}{\lambda}-\dfrac{t}{T} - \dfrac{x}{\lambda}+\dfrac{t}{T} \right]+\phi_{20}$$
$$\Delta \phi_{\rm Right}= \dfrac{ -2\pi L}{\lambda} +\phi_{20}$$
And since we need the interference in the right to be perfectly constructive,
$$\Delta \phi_{\rm Right}= \dfrac{ -2\pi L}{\lambda} +\phi_{20}=2\pi n\tag 1$$
$$D_{1,\rm Left}=a\sin\left[ 2\pi\left(\dfrac{-x}{\lambda}-\dfrac{t}{T}\right)\right]$$
$$D_{2,\rm Left}=a\sin\left[ 2\pi\left(\dfrac{-x+L}{\lambda}-\dfrac{t}{T}\right)+\phi_{20}\right]$$
Hence, the phase difference between them on the left side is given by
$$\Delta \phi_{\rm Left}=\phi_2-\phi_1$$
$$\Delta \phi_{\rm Right}= 2\pi\left(\dfrac{-x+L}{\lambda}-\dfrac{t}{T}\right)+\phi_{20}-\left[ 2\pi\left(\dfrac{-x}{\lambda}-\dfrac{t}{T}\right)\right]$$
$$\Delta \phi_{\rm Right}= 2\pi\left[\dfrac{-x+L}{\lambda}-\dfrac{t}{T} - \left(\dfrac{-x}{\lambda}-\dfrac{t}{T}\right)\right]+\phi_{20}$$
$$\Delta \phi_{\rm Right}= 2\pi\left[\dfrac{-x+L}{\lambda}-\dfrac{t}{T} - \dfrac{-x}{\lambda}+\dfrac{t}{T}\right]+\phi_{20}$$
$$\Delta \phi_{\rm Right}= 2\pi\left[\dfrac{L}{\lambda} \right]+\phi_{20}$$
And since we need the interference in the right to be perfectly destructive,
$$\Delta \phi_{\rm Right}= \dfrac{2\pi L}{\lambda} +\phi_{20}=2\pi\left(m+\frac{1}{2}\right)\tag 2$$
Solving (1) for $\phi_0$;
$$ \phi_{20}=2\pi n+ \dfrac{ 2\pi L}{\lambda}\tag 3$$
Plugging into (2);
$$ \dfrac{2\pi L}{\lambda} +2\pi n+ \dfrac{ 2\pi L}{\lambda}=2\pi\left(m+\frac{1}{2}\right)$$
$$ \dfrac{ L}{\lambda} + n+ \dfrac{ L}{\lambda}= m+\frac{1}{2} $$
The smallest value of $L$ occurs when $m=n$, so
$$ \dfrac{2 L}{\lambda}= \frac{1}{2} $$
$$\boxed{L= \frac{1}{4}\lambda}$$
$$\color{blue}{\bf [b]}$$
From (3);
$$ \phi_{20}=2\pi n+ \dfrac{ 2\pi L}{\lambda} $$
where $L= \frac{1}{4}\lambda$
$$ \phi_{20}=2\pi n+ \dfrac{ 2\pi ( \frac{1}{4}\lambda)}{\lambda} $$
$$ \phi_{20}=2\pi n+\frac{1}{2}\pi$$
Noting that the integer numbers of $2\pi$ is the same phase since it is one full circle, so $2\pi n$ does not change the phase.
So we can say that $n=0$, and hence,
$$ \boxed{ \phi_{20}= \frac{1}{2}\pi\;\rm rad}$$
$$\color{blue}{\bf [c]}$$
We know that $ \phi_{20}= \dfrac{\pi}{2}=\frac{1}{4}(2\pi)$, and $L=\frac{1}{4}\lambda$. So we can reach to that by delaying the wave by the same fraction of $T$,
$$\boxed{\Delta t=\frac{1}{4}T}$$
$$\color{blue}{\bf [d]}$$
$$\Delta t=\frac{1}{4}T$$
where $T=1/f$,
$$\Delta t=\frac{1}{4f}=\dfrac{1}{4\cdot 1000\times 10^2}$$
$$\Delta t=\color{red}{\bf 2.5\times 10^{-7}}\;\rm s$$
$$L=\frac{1}{4}\lambda$$
where $\lambda=c/f$
$$L=\frac{c}{4f}=\dfrac{3\times 10^8}{4\cdot 10^6}$$
$$L=\color{red}{\bf 75}\;\rm m$$
