As per the question it is given that,
Height/length of the cylindrical road roller $=h=1m=100cm$
Internal Diameter of the cylindrical road roller $=54cm$
Thus, the internal radius of the cylindrical road roller $=27cm=r$
It is given also, the thickness of the road roller (T) $=9cm$
Let us assume that the outer radii of the cylindrical road roller be R.
$T=R–r$
$9=R–27$
$R=27+9$
$R=36cm$
Then,
The volume of the iron sheet $\left( V \right)=\pi \times \left( {{R}^{2}}-{{r}^{2}} \right)\times h$
$V=\pi \times \left( {{36}^{2}}-{{27}^{2}} \right)\times 100$
$V=1780.38c{{m}^{3}}$
Therefore, the volume of the iron sheet $=1780.38c{{m}^{3}}$
Given also, mass of of the iron sheet $=7.8gm$
Thus, the mass of of the iron sheet $=1388696.4gm=1388.7kg$
Hence, the mass of the road roller is $1388.7kg$