If I understand your question, here is the solution.

The velocity of the car is 60 kph, at (90 -78) = 12 degrees above the East-axis.

The velocity of the bee in the car is 3m/sec, at (12 +37) = 49deg above the East-axis.

Km/h and m/sec.

Let use km/h all.

3m/sec *(1km /1000m)(3600sec /1hr) = 3*3600/1000 = 10.8 kph

So, Easting-components of the two velocities,

E = 60cos(12deg) +10.8cos(49deg) = 65.774 kph

Northing-components of the two velocities,

N = 60sin(12deg) +10.8sin(49deg) = 20.626 kph

Resultant of E and N,

R = sqrt[E^2 +N^2]

R = sqrt[(65.774)^2 +(20.626)^2] = 68.932 kph ---------***

Bearing, based from North,

tan(theta) = E/N = 65.774 /20.626 = 3.18888

theta = arctan(3.18888) = 72.589 degrees = 72deg and 35.34min --------***

Therefore, relative to the ground, the bee flies at the rate of 68.9 km/h on a bearing of 72deg,35min. ----------------answer.