Indeed, the plot can be obtained numerically.

T[R_?NumericQ] := First[r /. NSolve[R^2 (1 - 1/r Erf[r/4]) == r^2, r, Reals]] 
Potential[R_?NumericQ] := 1/R^2 (1 - 1/T[R] Erf[T[R]/4]) 
Plot[Potential[R], {R, 0, 5}]

[![enter image description here][1]][1]

Note that Potential is singular at R == 0. [1]: https://i.sstatic.net/xRCB1.png

Indeed, the plot can be obtained numerically.

T[R_?NumericQ] := First[r /. NSolve[R^2 (1 - 1/r Erf[r/4]) == r^2, r, Reals]] 
Potential[R_?NumericQ] := 1/R^2 (1 - 1/T[R] Erf[T[R]/4]) 
Plot[Potential[R], {R, 0, 5}, PlotRange -> {0, 5}]

Note that Potential is singular at R == 0, so it might be more informative to plot.

Plot[R^2 Potential[R], {R, 0, 5}]