Solve for w $($pillar width$)$

Factor $of$ safety$=\frac{\text{P}\text{i}\text{l}\text{l}\text{a}\text{r}\text{S}\text{t}\text{r}\text{e}\text{n}\text{g}\text{t}\text{h}}{\text{P}\text{i}\text{l}\text{l}\text{a}\text{r}\text{s}\text{t}\text{r}\text{e}\text{s}\text{s}}=1\mathrm{.}5$

Pillar Strength $=on(0\mathrm{.}778+0\mathrm{.}222*\frac{w}{H})$

wher ${}_n($ strength $of$ rock mass$)=24\mathrm{.}3$ MPa

$H($ extraction heigth$)=2\mathrm{.}5$m

Stress $=h(\frac{w+R}{w}{)}^2$

ahere $(unit$ weight ofoverburden$)=27$ KN$/$m$^3$

$h(pillar$ height$)=250m$

$R(depth$ below surface$)=10m$