Height of Cylinder $=h=AO=2.4\text{cm}$

Diameter of Cylinder $=1.4\text{cm}$

Radius of Cone = Radius of Cylinder
$OB=r=\frac{1.4}{2}=0.7\text{cm}$

Lets find Slant Height $l$ of cone, using pythagoras theorem on $△AOB$ , we get

$AB=l=\sqrt{h^2+r^2}=\sqrt{(2.4{)}^2+(0.7{)}^2}$
$l=\sqrt{5.76+0.49}=\sqrt{6.25}=2.5\text{cm}$

Surface Area of remaining Solid
= Surface Area of the Cylinder $+$ Inner surface Area of the hollow Cone 
          $=(2\pi rh+\pi r^2)+\pi rl=\pi r(2h+r+l)=\frac{22}{7}\times 0.7(2\times 2.4+0.7+2.5)=2.2(4.8+0.7+2.5)=2.2\left(8\right)=17.60{\text{cm}}^2$ 

$\therefore$ Total surface area of the remaining solid is $17.60{\text{cm}}^2$