Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The parallel rays that hit the left side of our lens system will experience no diffraction until it hits the first curved surface.
Now we have two curved surfaces. The rays reach the left side of it [from the flat-curved lens] parallel as we explained above, so $s=\infty$ and exit into the second lens.
The first image from the flat lens is given by
$$\dfrac{n_1}{s }+\dfrac{n_2}{s'}=\dfrac{n_2-n_1}{R}$$
$$\dfrac{n_1}{\infty}+\dfrac{n_2}{s'}=\dfrac{n_2-n_1}{R}$$
Hence,
$$s'=\dfrac{n_2R}{n_2-n_1}\tag 1$$
Now the rays leave the second lens to air.
So,
$$\dfrac{n_2}{s_1 }+\dfrac{n_{air}}{s_1'}=\dfrac{n_{air}-n_2}{R}$$
where $s_1=s'$ since the two lenses are in touch and the distance between them is zero. So plug from (1),
$$\dfrac{n_2}{\dfrac{n_2R}{n_2-n_1}}+\dfrac{n_{air}}{s_1'}=\dfrac{n_{air}-n_2}{R}$$
$$\dfrac{n_2-n_1}{R}+\dfrac{n_{air}}{s_1'}=\dfrac{n_{air}-n_2}{R}$$
Noting that $s_1'=-f$ since the rays will be gathered by the two-lens system to a focus point $f_{\rm effective}=f$ and the negative sign is due to the sign rules (the right side of the system is concave toward the first image $s'$).
$$\dfrac{n_2-n_1}{R}+\dfrac{n_{air}}{-f}=\dfrac{n_{air}-n_2}{R}$$
Solving for $f$,
$$\dfrac{n_2-n_1}{R}-\dfrac{n_{air}-n_2}{R}=\dfrac{n_{air}}{f} $$
where $n_{air}=1$,
$$\dfrac{n_2-n_1}{R}-\dfrac{1-n_2}{R}=\dfrac{1}{f} $$
$$ \dfrac{n_2-n_1-1+n_2}{R}=\dfrac{1}{f} $$
Therefore,
$$\boxed{f=\dfrac{R}{2n_2-n_1-1}}$$
$$\color{blue}{\bf [b]}$$
For red light, using the boxed formula above,
$$f_{\rm red}=\dfrac{R}{2n_{2,\rm red}-n_{1,\rm red}-1}$$
And for blue light, using the boxed formula above,
$$f_{\rm blue }=\dfrac{R}{2n_{2,\rm blue }-n_{1,\rm blue }-1}$$
And when $f_{\rm red}=f_{\rm blue}$, then the two equation above are equal. So that
$$\dfrac{ \color{red}{\bf\not} R}{2n_{2,\rm red}-n_{1,\rm red}-1}=\dfrac{ \color{red}{\bf\not} R}{2n_{2,\rm blue }-n_{1,\rm blue }-1}$$
Thus,
$$ 2n_{2,\rm red}-n_{1,\rm red}- \color{red}{\bf\not} 1=2n_{2,\rm blue }-n_{1,\rm blue }- \color{red}{\bf\not} 1$$
$$ 2n_{2,\rm red}-n_{1,\rm red} =2n_{2,\rm blue }-n_{1,\rm blue } $$
We need to find $\Delta n_2$, which is given by $n_{2,\rm blue}-n_{2,\rm red}$,
$$n_{1,\rm blue } -n_{1,\rm red} =2n_{2,\rm blue }- 2n_{2,\rm red} $$
Thus,
$$\Delta n_1=2 \Delta n_2 $$
$$\boxed{\Delta n_2=\dfrac{\Delta n_1}{2}}$$
$$\color{blue}{\bf [c]}$$
To find which is which, we need to find $\Delta n_1$ and $\Delta n_2$ for both cases. The one that have one is twice the other is our choice for $n_1$.
$$\Delta n_{\rm Crown \;glass}=n_{blue}-n_{red}=1.525- 1.517=\bf 0.008 $$
$$\Delta n_{\rm flint\;glass}=n_{blue}-n_{red}=1.632- 1.616=\bf 0.016 $$
Hence,
$$\Delta n_{\rm Crown \;glass}=\dfrac{\Delta n_{\rm flint\;glass}}{2}$$
which is similar to $\Delta n_2=\dfrac{\Delta n_1}{2}$.
Hence,
The flint glass will be for the diverging glass and the crown glass will be for the converging glass.
$$\color{blue}{\bf [d]}$$
Solving the first boxed formula above for $R$,
$$R= f(2n_2-n_1-1) $$
And we know that $f_{\rm red}=f_{\rm blue}$, so we can use any color.
Let's take the red one.
Noting that $n_{2}=n_{\rm Crown \;glass }$, and $n_1=n_{\rm flint \;glass}$,
$$R= 10(2[1.517]-1.616-1) =\color{red}{\bf 4.18}\;\rm cm $$
