According to the question

${\mathrm{n}}_1=20, {\mathrm{n}}_2=45, {\mathrm{n}}_3=15, {\mathrm{a}}_1=10, {\mathrm{a}}_2=5, {\mathrm{a}}_3=3$

We know the formula,

Weighted average(Aw)$=\frac{{\mathrm{n}}_1{\mathrm{a}}_1+{\mathrm{n}}_2{\mathrm{a}}_2+{\mathrm{n}}_3{\mathrm{a}}_3}{{\mathrm{n}}_1+{\mathrm{n}}_2+{\mathrm{n}}_3}$

Average price of all the stationary goods

$$\begin{gathered} =\frac{20\times 10+45\times 5+15\times 3}{20+45+15} \\ =\frac{200+225+45}{80} \\ =\frac{470}{80} \end{gathered}$$

= 5.875 yr

Hence, Option A is correct.