The reaction is
[tex]U(s)+3F_{2} (g)-->UF_{6} (s)[/tex]
the Q of reaction will be
[tex]Q = \frac{1}{(pF_{2})^{3}}[/tex]
[tex]Q = \frac{1}{(0.054)^{3} } = 6350.66[/tex]
Δ[tex]G=[/tex]Δ[tex]G^{0}+RTlnQ[/tex]
Putting values
ΔG = -2068000 + (8.314)(298)lnQ
ΔG = -2068000 + (8.314)(298)ln(6350.66) = -2046305 J /mol = -2046.3 kJ / mol
