Number of neutrons present in ${}_{14}\mathrm{Si}^{28}$ is twice the number of protons present in the element ${\mathrm{X}}^{14}$. Then ${\mathrm{X}}^{14}$ and ${}_6\mathrm{C}^{13}$ are

Let number of protons in X¹⁴ = p

number of neutrons in Si = 28 - 14 = 14

2p = 14 

p = 7

Number of neutrons in ${}_7\mathrm{X}^{14}$ = 14 - 7 = 7

Number of neutrons in ${}_6\mathrm{C}^{13}$ = 13 - 6 = 7