The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm$^{-3}$, respectively. If the standard free energy difference ($\Delta G^\circ$) is equal to 1895 J mol$^{-1}$, the pressure at which graphite will be transformed into diamond at 298 K is -

Graphite changes into diamond when the work done by the applied pressure on the decrease in volume supplies the free energy difference: $\Delta G^\circ = P\,|\Delta V|$.
Molar volume of graphite $= \dfrac{12}{2.25} = 5.33$ cm$^3$ mol$^{-1}$
Molar volume of diamond $= \dfrac{12}{3.31} = 3.63$ cm$^3$ mol$^{-1}$
$|\Delta V| = 5.33 - 3.63 = 1.71$ cm$^3$ mol$^{-1}$ $= 1.71 \times 10^{-6}$ m$^3$ mol$^{-1}$
$P = \dfrac{\Delta G^\circ}{|\Delta V|} = \dfrac{1895}{1.71 \times 10^{-6}} \approx 1.1 \times 10^9$ Pa
This is of the order of $10^9$ Pa, i.e. about $10^4$ atmospheres. Among the options, the value of this order is $9.92 \times 10^8$ Pa; the other options are 10, 100 or 1000 times too small.
So graphite will be transformed into diamond at about $9.92 \times 10^8$ Pa.