Would be easier with a diagram but here it goes

Let the triangle be ABC

Let the circle be DEF (the 3 points of contact) (D lies on AB, E on BC, F on AC)

O will be the centre

Join OD , OE, OF and OB , OA , OC

In Triangle OBE and Triangle OBD

OD is perpendicular to AB and OE is perpendicular to BC, So 90^{0}

OE = OD (Radii)

OB = OB (Common)

Therefore Triangle OBE is congruent to Triangle OBD

Hence, Angle OBE = Angle OBD

Similarly, in the other triangles , we get

Angle OCE = Angle OCA

Angle OAF = Angle OAD

Now Angle CAB, ABC, BCA = 60 degrees (eq. triangle)

Which means Angle OBD = OBE = 30 degrees (half)

Do Tan (Angle OBE)

Tan30 = 1/√3

1/√3 = radius (2 cm) / BE

Therefore BE = 2√3

Similarly, CE = 2√3

So BC = 4√3

Ar(eq. Triangle) = [(Side) ^{2} √3] / 4

= 48√3 / 4

= 12√3 cm^{2}

Answer is lengthy coz of that Angle, just keep in mind that when you join OA, OB and OC, Angle A, B and C are bisected, the congruence shown above is its proof.

:)