It is very easy.    First make a sketch  to follow my explanations.


You have DB as a tangent line to the circle released from the point D outside the circle..


You have DC as a secant line to the circle released from the point D outside the circle.


Now, you should use this property 


    If a tangent and a secant lines are released from a point outside a circle,  then the product of the measures 
    of the secant and its external part is equal to the square of the tangent segment. 



It is well known property of tangent segments and secants.
See the lesson  Metric relations for a tangent and a secant lines released from a point outside a circle in this site.


In your case you need to find the altitude BD, which is the tangent to the circle, as it is given.


Your secant DC has the total length of 48+12 = 60 cm.
Its external part is 12 cm long.


So,  |DB|^2 = |DC|*|DA| = 12*60 = 720,

hence  |DB| =  = 26.83 cm.    Option d)