If p is an integral multiple of 4 lying in between the coefficients of ${\mathrm{x}}^4$ and x in the expansion of ${\left({\mathrm{x}}^2+\frac{1}{\mathrm{x}}\right)}^8,$ then the number of such values of p is

Coeff  ${\mathrm{x}}^4, \mathrm{r}=\frac{8\left(2\right)-4}{3}=\frac{12}{3}=4$

Coeff $\mathrm{x}, \mathrm{r}=\frac{8\left(2\right)-1}{3}=5$

                $$\begin{gathered} {}^8{\mathrm{C}}_5<\mathrm{P}<{}^8{\mathrm{C}}_4 \\ 56<\mathrm{P}<70 \end{gathered}$$

Multiples of 4 in (56, 70)

       = 60, 64, 68

Required number = 3