Let the original radius of the base be r and the original height be h.
From the given condition:

h = 2r

The volume of a cylinder is given by:

V = π r² h

Substituting h = 2r in the volume formula:

V₁ = π r² (2r)

V₁ = 2π r³

Now, the new height is 6 times the radius, so:

h’ = 6r

The new volume is:

V₂ = π r² (6r)

V₂ = 6π r³

According to the problem, the difference in volume is 539 cubic decimeters:

V₂ – V₁ = 539

Substituting the values of V₁ and V₂:

6π r³ – 2π r³ = 539

4π r³ = 539

Substituting π ≈ \frac{22}{7} :

4 × \frac{22}{7} × r³ = 539

\frac{88}{7} × r³ = 539

Multiplying both sides by 7:

88 r³ = 3773

⇒ r³ = \frac{3773}{88}

⇒ r³ = \frac{343}{8}

⇒ r = 3.5 dm

∴ height = 2r

= 2 × 3.5

= 7 cm