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 $1388.7kg$