The volumes of $4\mathrm{N} \mathrm{HCl}$ and $10\mathrm{N} \mathrm{HCl}$ required to make $1 \mathrm{L}$ of $6\mathrm{N} \mathrm{HCl}$ are

Total volume $\left(\mathrm{V}\right) =1 \mathrm{L}$(given)

Normality $\left(\mathrm{N}\right)=6 \mathrm{N}$(given)

Let,

-${\mathrm{N}}_1$ be normality of $4\mathrm{N} \mathrm{HCl}; {\mathrm{V}}_1$ is the volume of $4\mathrm{N} \mathrm{HCl}$

- ${\mathrm{N}}_2$ be the normality of $10\mathrm{N} \mathrm{HCl}; {\mathrm{V}}_2$ be the volume of $10\mathrm{N} \mathrm{HCl}$

- Now, ${\mathrm{V}}_1+{\mathrm{V}}_2=\mathrm{V}$

Therefore, ${\mathrm{V}}_2=\mathrm{V}-{\mathrm{V}}_1$

- We know that

According to the neutralization law,

 ${\mathrm{N}}_1{\mathrm{V}}_1+{\mathrm{N}}_2{\mathrm{V}}_2=\mathrm{NV}$

- Substituting the values in the above formula we,

$4{\mathrm{V}}_1+10\left(1-{\mathrm{V}}_1\right)=6\times 1$

therefore${\mathrm{V}}_1=\frac{2}{3}$

- Similarly, 

${\mathrm{V}}_2=1-\frac{2}{3}=\frac{1}{3}$

Thus, ${\mathrm{V}}_1=0.67 \mathrm{L}$  and ${\mathrm{V}}_2=0.33\mathrm{L}$ 

Hence, option(c) is the correct answer.